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Category: Algebra

Question-144031

Question Number 144031 by mathdanisur last updated on 20/Jun/21 Answered by mindispower last updated on 20/Jun/21 1sin(2k)+1tg(2k)=1+cos(2k)sin(2k)=1tg(2k1)$$\frac{\mathrm{1}}{{sin}\left(\mathrm{2}^{{k}} \right)}=\frac{\mathrm{1}}{{tg}\left(\mathrm{2}^{{k}−\mathrm{1}} \right)}−\frac{\mathrm{1}}{{tg}\left(\mathrm{2}^{{k}} \right)}…